Take the mean distance of the moon and the sun from the earth to be 0.4× 10^6 km… - JEE Main 2018 Physics PYQ

JEE Main 2018 — Physics Mechanics

2018mcqhard

Take the mean distance of the moon and the sun from the earth to be $0.4\times {10}^{6}\mathrm{km}$ and $150\times {10}^{6}\mathrm{km}$ , respectively. Their masses are $8\times {10}^{22}$ $\mathrm{kg}$ and $2\times {10}^{30}$ $\mathrm{kg}$ , respectively. The radius of the earth is $6400\mathrm{km}$ . Let $\Delta {F}_{1}$ be the difference in the forces exerted by the moon at the nearest and farthest point on the earth, and $\Delta {F}_{2}$ be the difference in the forces exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $\frac{\Delta {F}_{1}}{\Delta {F}_{2}}$ is,

Official previous-year question

Held on 15 Apr 2018 · Verified 6 Jul 2026.

Options

A
$6$
B
${10}^{-2}$
C
$2$
D
$0.6$

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Solution

${F}_{1}=\frac{G{M}_{e}m}{{r}_{1}^{3}},{F}_{2}=\frac{G{M}_{e}{M}_{s}}{{r}_{2}^{2}}$

$\Delta {F}_{1}= (-)\frac{2G{M}_{e}m}{{r}_{1}^{3}} \Delta {r}_{1} , \Delta {F}_{2}=(-)\frac{2G{M}_{e}{M}_{s}}{{r}_{2}^{2}} \Delta {r}_{2}$

$\frac{\Delta {F}_{1}}{\Delta {F}_{2}}= \frac{m\Delta {r}_{1}}{{r}_{1}^{3}} \frac{{r}_{2}^{3}}{{M}_{s}\Delta {r}_{2}}=(\frac{m}{{M}_{s}})(\frac{{r}_{2}^{3}}{{r}_{1}^{3}})(\frac{\Delta {r}_{1}}{\Delta {r}_{2}})$

Using $\Delta {r}_{1}=\Delta {r}_{2}=2{R}_{earth}$

$m=8\times {10}^{22} \mathrm{kg}$

$Ms= 2\times {10}^{30} \mathrm{kg}$

${r}_{1}=0.4\times {10}^{6}\mathrm{km}$

${r}_{2}=150\times {10}^{6}\mathrm{km}$

We get $\frac{\Delta {F}_{1}}{\Delta {F}_{2}}=2$ .

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